Solve the following equation for k 2k + 1 ÷ 6 + 4k = 6k - 5÷ 9 -4
Answers
Answer:
step
1
:
equation at the end of step 1
(((((3•(k2))-2k)-1)•(-k-1))-((4k-2)•(((k2)-6k)+1)))+((k+3)•(3k+1)•(k-3)) = 0
step
2
:
equation at the end of step 2
(((((3•(k2))-2k)-1)•(-k-1))-((4k-2)•(((k2)-6k)+1)))+(k+3)•(3k+1)•(k-3) = 0
step
3
:
step
4
:
pulling out like terms
4.1 pull out like factors :
4k - 2 = 2 • (2k - 1)
trying to factor by splitting the middle term
4.2 factoring k2 - 6k + 1
the first term is, k2 its coefficient is 1 .
the middle term is, -6k its coefficient is -6 .
the last term, "the constant", is +1
step-1 : multiply the coefficient of the first term by the constant 1 • 1 = 1
step-2 : find two factors of 1 whose sum equals the coefficient of the middle term, which is -6 .
-1 + -1 = -2
1 + 1 = 2
observation : no two such factors can be found !!
conclusion : trinomial can not be factored
equation at the end of step
4
:
(((((3•(k2))-2k)-1)•(-k-1))-2•(2k-1)•(k2-6k+1))+(k+3)•(3k+1)•(k-3) = 0
step
5
:
equation at the end of step
5
:
((((3k2-2k)-1)•(-k-1))-2•(2k-1)•(k2-6k+1))+(k+3)•(3k+1)•(k-3) = 0
step
6
:
step
7
:
pulling out like terms
7.1 pull out like factors :
-k - 1 = -1 • (k + 1)
trying to factor by splitting the middle term
According to the Quadratic Formula, k , the solution for Ak2+Bk+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
k = ————————
2A
In our case, A = 3
B = -2
C = -1
Accordingly, B2 - 4AC =
4 - (-12) =
16
Applying the quadratic formula :
2 ± √ 16
k = —————
6
Answer:
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